Let $H$ be a Hilbert space and $T \in L(H)$ a bounded linear operator.If $\exists c>0$ such that $Re(\langle Tx,x \rangle) \geqslant c||x||^2, \forall x \in H$, then prove that $\text{Ran}(T^*)=H$ where $T^*$ is the adjoint operator of $T$.
This exercise was given by my instructor and he gave us a hint to prove that $\text{Ran}(T^*)$ is a closed subspace of $H$
Proof:
It's very easy to prove that $\text{Ran}(T^*)$ is a subspace of $H$.
Let $x_n \in \text{Ran}(T^*)$ such that $x_n \longrightarrow x$, then $x_n=T^*z_n$.We have that $T^*z_n$ is a Cauchy sequence, thus $\exists n_0 \in \mathbb{N}$ such that $||T^*z_n-T^*z_m||< c\epsilon, \forall m,n \geqslant n_0$
Now $$||z_n-z_m||^2 \leqslant \frac{1}{c} Re\langle T(z_n-z_m),z_n-z_m \rangle\leqslant \frac{1}{c}| \langle T(z_n-z_m),z_n-z_m\rangle|= \frac{1}{c}|\langle z_n-z_m,T^*(z_n-z_m)\rangle| \leqslant \frac{1}{c}(||z_n-z_m||)||T^*z_n-T^*z_m||$$ so $$||z_n-z_m|| \leqslant \frac{1}{c}||T^*z_n-T^*z_m||< \epsilon$$
But $H$ is a Hilbert space hence complete, thus $z_n \longrightarrow z$ and $T^*z_n \longrightarrow Tz$ and also $T^*z_n \longrightarrow x \Longrightarrow x=T^*z \Longrightarrow x \in \text{Ran}(T^*)$ thus $\text{Ran}(T^*)$ is a closed subspace of $H$ .
Also $\text{Ran}(T^*)$ is complete as a closed subspace of a complete space.
Clearly $\text{Ran}(T^*) \subseteq H$
Now let $x \in H$.
Because of the fact that $\text{Ran}(T^*)$ is a complete subspace of $H$ we have that exists a unique $y_0 \in R(T^*)$ such that $\inf\{||x-y||:y \in R(T^*)\}= ||x-y_0||\}$ and $x-y_0 \perp R(T^*)$
Thus $||x-y_0||^2 \leqslant \frac{1}{c}Re \langle x-y_0,T^*(x-y_0) \rangle=0 \Longrightarrow \inf\{||x-y||:y \in R(T^*)\}=||x-y_0||=0\Longrightarrow x \in \text{Ran}(T^*)$ because $\text{Ran}(T^*)$ is a closed subspace of $H$, thus $H=\text{Ran}(T^*)$
Does my proof have any error?
Thank you in advance.